I’ll throw in some coordinate-free-ness, for originality’s sake, and because I hate matrices.

Let \(V\) be a complex vector space and let \(\Lambda\) be a finitely generated torsion-free Abelian group such that \(\Lambda \otimes {\mathbb{R}}\cong V\) as \({\mathbb{R}}\)-vector spaces. The data of \(V\), \(\Lambda\) and a Hermitian form \(H: V \otimes V \to {\mathbb{C}}\) such that \(H|_{\Lambda \times \Lambda} \subset {\mathbb{Z}}\) gives rise to an Abelian variety and a line bundle endowed with a Hermitian metric.

Let \(A \in GL(\Lambda)\) be an intergral linear invertible self-map such that \((Ax, y) = -(x,Ay)\) wrt the standard inner product on \(\Lambda\). Via \({\mathbb{R}}\)-linear extension this induces a symplectic form on \(V\), call it \(\omega\). Define the following Hermitian metric on \(V\)\[
H(x,y) = \omega(Ix,y) + i \omega(x,y)
\] where \(I\) is the complex structure on \(V\). One checks that it is \({\mathbb{C}}\)-linear in the first argument, \({\mathbb{C}}\)-antilinear in the second. Further, \[
H(x,y) = \overline{H(y,x)} \qquad \textrm{ iff }\qquad \omega(Ix,Iy) = \omega(x,y)
\]

Since \(\Pi\) is an isomorphism and in particular surjective, in order to check that \(H\) is Hermitian we need to check that \(\omega(I \Pi x, I \Pi y) = \omega(\Pi x, \Pi y)\) for all \(x,y \in \Lambda \otimes {\mathbb{R}}\). Remark now that from definition of \(\omega\), \(\omega(\Pi x, \Pi y)=(Ax,y)\).

Pass to the complexification: let \(\Pi \otimes {\mathbb{C}}: (\Lambda \otimes {\mathbb{R}}) \otimes_{{\mathbb{R}}} C \to V \otimes_{{\mathbb{R}}} {\mathbb{C}}\) be the \({\mathbb{C}}\)-linear extension of \(\Pi\). The vector space \(V\) decomposes into direct sum of eigenspaces of \(I\): \(V = V^{1,0} \oplus V^{0,1}\) where \(I\) acts by \(i\) on the first piece and by \(-i\) on the second. Denote \(\Pi' := \pi_{V^{1,0}} \circ \Pi \otimes {\mathbb{C}}\). By definition, \(\Pi'\) is the same as the composition of \(\Pi\) and identification between \(V \subset V \otimes C\) and \(V^{1,0}\) via the projection on \(V^{1,0}\). When written out in some basis \(\Pi'\) is called a period matrix. If \(\Lambda \cong {\mathbb{Z}}^{2g}\), it is a \(g \times 2g\) matrix which has the coordinates of the generators of \({\mathbb{Z}}^{2g}\) as its columns.

After restriction to \(V^{1,0}\) it is clear that \(\omega\) respects \(I\) if and only if \(\omega_{\mathbb{C}}(i \Pi' x, i \Pi'y) = \omega_{\mathbb{C}}(\Pi' x, \Pi'y)\) for all \(x,y \in \Lambda \otimes C\).

Define \[
J = (\Pi\otimes {\mathbb{C}})^{-1} (I\otimes {\mathbb{C}}) (\Pi\otimes {\mathbb{C}})
\] Notice that it has the property \(i \Pi' = \Pi' J\). Also notice that \(\omega_{\mathbb{C}}(\Pi' x, \Pi' y) = (Ax, y)\). Then the condition to check becomes \[
(A J x, J y) = (A x, y)
\] or \(J^* A J=A\). Rewrite the latter identity using the definition of \(J\), denoting \(\Phi = \Pi \otimes {\mathbb{C}}\)\[
\Phi^* (I\otimes {\mathbb{C}})^* (\Phi^{-1})^* (A \otimes {\mathbb{C}}) \Phi^{-1} (I\otimes {\mathbb{C}}) \Phi = A
\] so, using self-adjointness of \(I \otimes {\mathbb{C}}\), \[
I\otimes {\mathbb{C}}(\Phi^{-1})^* (A \otimes {\mathbb{C}}) \Phi^{-1} (I\otimes {\mathbb{C}}) = (\Phi^{-1})^* A \Phi^{-1}
\] and further (omitting \(\otimes {\mathbb{C}}\) for simplicity of notation) \[
I (\Phi A^{-1} \Phi^*)^{-1} I = (\Phi A^{-1} \Phi^*)^{-1}
\] Given that \(I \otimes {\mathbb{C}}\) is semi-simple, the restriction of \((\Phi A^{-1} \Phi^*)^{-1}\) to \(V^{1,0}\) must take values in \(V^{0,1}\) (if the projection of \((\Phi A^{-1} \Phi^*)^{-1}(x)\) on \(V^{1,0}\) is non-zero element \(y\), then \(i y i = y\) which is absurd), and similar reasoning works for \(V^{0,1}\). We conclude, computing \(\pi_{1,0} \circ (\Phi A^{-1} \Phi^*)^{-1}\) and \(\pi_{0,1} \circ (\Phi A^{-1} \Phi^*)^{-1}\), that \[
\Pi' A^{-1} (\Pi')^* = 0 \textrm{ and } \overline{\Pi'} A^{-1} (\overline{\Pi'})^*
\] [the trick is to write down the operator to be inverted as an inverse of block matrix, and observe that the “determinant” term does not matter for vanishing] The second equality is the conjugation of the first, and we are done clarifing the condition for \(H\) to be Hermitian.

Now let us compute \(H(x,x)\). As before it will be convenient to pass to the complexification of \(\omega\) and put \[
H_{\mathbb{C}}(x,y) = \omega_{\mathbb{C}}(I \otimes {\mathbb{C}}x, y) +i \omega_{\mathbb{C}}(x,y)
\] The second term is defined by the proprety: \(\omega_{\mathbb{C}}(\Phi x, \Phi y) = (Ax,y)\), and hence \[
\omega_{\mathbb{C}}(x, y) = (A\Phi^{-1}x,\Phi^{-1}y)
\] so \[
\omega_{\mathbb{C}}(x, y) = ((\Phi^{-1})^* A \Phi^{-1}x, y) = ((\Phi A
\Phi^*)^{-1} x, y)
\] And so the first term is: \[
\omega_{\mathbb{C}}(I \otimes {\mathbb{C}}x, y) = ((\Phi^* A
\Phi)^{-1} (I \otimes {\mathbb{C}})x, y)
\] Assuming \(H\) to be Hermitian, the restriction of \((\Phi^* A^{-1} \Phi)^{-1}\) to \(V^{1,0}\) writes \((\Pi' A^{-1} (\overline{\Pi'})^*)^{-1} ) D\), and its restriction to \(V^{0,1}\) is \((\overline{\Pi'} A^{-1} (\Pi')^*)^{-1}\). Therefore \(H(x,x)\) is \[
(2i (\Pi' A^{-1} (\overline{\Pi'})^*)^{-1} x,x)
\] on \(V^{1,0}\). Since positivity depends only on the sign of eigenvalues, passing to the inverse operator does not affect it, and so the condition for positivity of \(H\) is \[
(i (\Pi' A^{-1} (\overline{\Pi'})^*)^{-1} x,x) > 0
\] for all \(x \in V \hookrightarrow V \otimes {\mathbb{C}}\).